Question

Find the sum of $n$ terms of the series whose $n^{th}$ term is
$n(n+2)$.

Answer 1

The general term is given by,

$T_n=n(n+2)=n^2+2n$

Now, $\Sigma T_n=\Sigma (n^2+2n)$

$\Sigma T_n=\Sigma n^2+\Sigma 2n$

$\Sigma T_n=\Sigma n^2+2\Sigma n$ ------------- (1)

We know that,

$\Sigma n^2=\frac{n(n+1)(2n+1)}{6}$

and $\Sigma n=\frac{n(n+1)}{2}$

Using these summation in equation (1) ,

$\Sigma T_n=\frac{n(n+1)(2n+1)}{6}+\frac{2n(n+1)}{2}$

$=\frac{2n^3+3n^2+n}{6}+n^2+n$

$=\frac{2n^3+3n^2+n+6n^2+6n}{6}$

$=\frac{2n^3+9n^2+7n}{6}$

Hence, $\Sigma T_n=\frac{2n^3+9n^2+7n}{6}$